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Lemma 2 Let $f: X \rightarrow Y$ be a morphism. Then $f$ is separated if $\Delta(X)$ is closed in $X \times_Y X$.

Proof: Since we know that $\Delta(X)$ is closed, what is left to show is that the morphism of sheaves $\Delta: \mathcal{O}_{X \times_Y X} \rightarrow \Delta^* \mathcal{O}_X$ is surjective. Suffices to check it locally, in a neighbourhood of any point of $\Delta(X)$. But this follows from the affine case. In particular, we can cover $X$ and $Y$ by open affines, $X_i$ and $Y_j$ respectively, such that for any $X_i$ there is a $Y_j$ such that $f(X_i) \subset Y_j$ and apply the previous proposition to $X_i \times_{Y_j} X_i$ (indeed, these sets cover $X \times_Y X$). ☐

A subring $R$ of a field $K$ is called a valuation ring if for any $x \in K$ either $x \in R$ or $x^{-1} \in R$. Then $\mathfrak{m} = R \setminus R^\times \subset R$ is the maximal ideal of $R$. The factor $\Gamma = K^\times / R^\times$ is called the valuation group and the natural projection on the factor, $v: K^\times \rightarrow \Gamma$, is called the valuation map. By declaring the elements of $\mathfrak{m}$ positive one turns $\Gamma$ into a (totally) ordered group.

There is a correspondence

prime ideals of $R$

$\Longleftrightarrow$

subgroups $\Gamma'$ of $\Gamma$ such that $0 < y \leq x, x \in \Gamma'$ implies $y \in \Gamma'$

 

right to left is given by

$\displaystyle I(\Gamma') = \{ x \in R^\times \mid v(x) \notin \Gamma' \} $

In particular, if $\Gamma = \mathbb R$, $R$ has only one non-trivial prime ideal, $\mathfrak m$. We will further consider only valuation ring of this kind, i.e. $\mathrm{Spec}~R = \{(0), \mathfrak{m}\}$.

Let $X$ be a topological space. The specialisation pre-order on $X$ is defined as

$\displaystyle x \leq y \textrm{ iff } \overline{\{x\}} \subset \overline{\{y\}} $

If $X$ satisfies the $T_0$ separation axiom (for any $x,y \in X$ there is an open $O_x \ni x$, $y \notin O_x$ or there is an open $O_y \ni y, x \notin O_y$) then this is an order. If $x \leq y$ then one says that $x$ is a specilisation of $y$.

Theorem 3 (Valuative criterion of separatedness) Let $f: X \rightarrow Y$ be a morphism of schemes, where $X$ is Noetherian. The morphism $f$ is separated if and only if for any valuation ring $R \subset K$ and any commuting diagram

where $\iota^*$ is the morphism of schemes induced by the inclusion $R \subset K$, there is at most one morphism $\mathrm{Spec}~R \rightarrow X$ completing the diagram

This breaks down into several lemmas. 

Lemma 4 Let $R \subset K$ be a valuation ring, $\mathrm{Spec}~R=\{\mathrm{Spec}~k, \mathrm{Spec}~K\}$. Let $\iota: \mathrm{Spec}\,k \rightarrow \mathrm{Spec}~R, \iota': \mathrm{Spec}~K \rightarrow \mathrm{Spec}~R $ be the natural morphisms. Let $f_1, f_2: \mathrm{Spec}~R \rightarrow X$ be two morphisms such that

$\displaystyle \begin{array}{lll} f_1 \circ \iota & = & f_1 \circ \iota\\ f_1 \circ \iota' & = & f_2 \circ \iota'\\ \end{array} $

Then $f_1 = f_2$.  

Proof: Let $U=\mathrm{Spec}{}A$ be an open affine neighbourhood of $\iota(\mathrm{Spec}~k) \in X$. Since $f_1, f_2$ are continuous they factor through $U$ therefore we have maps $f_1^*: A \rightarrow R \hookrightarrow K, f_2^*: A \rightarrow R \hookrightarrow K$ that coincide as morphisms to $K$ (since $f_1 \circ \iota' = f_2 \circ \iota'$), but then they just coincide. ☐

Let $R, S$ be two local subrings in $K$ with maximal ideals $\mathfrak{m}, \mathfrak{n}$. Then $R$ dominates $S$ if $R \supset S$ and $\mathfrak{m} \cap S = \mathfrak{n}$.

The stronger version à la Hartshorne:

Lemma 5 Let $R \subset K$ be a valuation ring. To give a morphism $f: \mathrm{Spec}{}R \rightarrow X$ is the same as to give the following data:

• - $x_0, x_1 \in X$, $x_0 \in \overline{\{x_1\}}$;
• - inclusion $\kappa(x_1) \subset K$ such that $R$ dominates $\mathcal{O}_{\overline{\{x_1\}}, x_0}$. 

Proof: Let $f: \mathrm{Spec}{}R \rightarrow X$ be a morphism. Let $x_0 = f(\mathrm{Spec}{}k)$, $x_1 = f(\mathrm{Spec}{}K)$. Take an affine $U \ni x$, $U = \mathrm{Spec}{}A$. Then the restriction of $f$ to $\mathrm{Spec}{}K$ gives a morphism $A \rightarrow K$. Let $\mathfrak{p}_0$ be the ideal of $A$ corresponding to $x_0$, and $\mathfrak{p}_1$ be the ideal corresponding to $x_1$, $\mathfrak{p}_1 \subset \mathfrak{p}_0$.

 Remarks:

• $(A/\mathfrak{p}_1) \otimes A_{\mathfrak{p}_1} = A_{\mathfrak{p}_1} / \mathfrak{p}_1 A_{\mathfrak{p}_1} = Q(A/\mathfrak{p}_1)$;
• $A/\mathfrak{p}_1 \rightarrow R$ is an inclusion, and since $A/\mathfrak{p}_1$ is an integral domain, $A/\mathfrak{p}_1 \rightarrow (A/\mathfrak{p}_1)_{\mathfrak{p}_0}$ is an inclusion too;
• we know that $\mathfrak{m} \cap A/\mathfrak{p}_1 = \mathfrak{p}_0$, it follows by tensoring with $(A/\mathfrak{p}_1)_{\mathfrak{p}_0}$ that $\mathfrak{m} \cap (A/\mathfrak{p}_1)_{\mathfrak{p}_0} = \mathfrak{p}_0 \otimes (A/\mathfrak{p}_1)_{\mathfrak{p}_0} = \mathfrak{m}_{x_0}$
• $(A/\mathfrak{p}_1)_{\mathfrak{p}_0} \subset R$. Indeed, an element from the first ring is of the form $a/b$ where $a \in A_{\mathfrak{p}_1}, b \notin \mathfrak{p}_0$. Since, $a \in R$, the question boils down to whether $b$ is invertible in $R$. But it is, since it does not belong to the maximal ideal;

Therefore, $R$ dominates over $\mathcal{O}_{\overline{x_1},x_0}$.

Given the data as in the statement of the lemma, choose an affine $U = \mathrm{Spec}{}A$, $x_0 \in U$. We are given the inclusion $\kappa(x_1) = Q(A/\mathfrak{p}_1) \rightarrow K$, therefore we have a morphism $A \rightarrow K$. We need to prove that it factors through $R$ and that the preimage of $\mathfrak{m} \subset R$ is $\mathfrak{p}_0 \subset A$, the ideal corresponding to $x_0$.

Since $R$ dominates $(A/\mathfrak{p}_1)_{\mathfrak{p}_0}$, $R \supset (A/\mathfrak{p}_1)_{\mathfrak{p}_0} \supset A/\mathfrak{p}_1$, and therefore our morphism goes actually to $R$: $f: A \rightarrow R$. The other part of the definition of domination tells us that $\mathfrak{m} \cap (A/\mathfrak{p}_1)_{\mathfrak{p}_0} = \mathfrak{p}_0 \otimes (A/\mathfrak{p}_1)_{\mathfrak{p}_0}$. Taking intersection with $A/\mathfrak{p}_1$ we get $\mathfrak{m} \cap A/\mathfrak{p}_1 = \mathfrak{p}_0 \otimes A/\mathfrak{p}_1$, which was to be shown.

☐

Lemma 6 (EGA II.7.2.1, essentially uses EGA I.6.6.5) Let $f: X \rightarrow Y$ be a quasi-compact morphism. Then $f(X)$ is closed in $Y$ if and only if $f(X)$ is stable under specialisations (i.e. $x \leq y$, $y \in f(X)$ implies $x \in f(X)$).

Proof: Given $f: X \rightarrow Y$ there exists $f': X_{red} \rightarrow Y_{red}$ that commutes with the natural projections, so we might as well suppose $X$ and $Y$ reduced. We also can replace $Y$ by $\overline{f(X)}$. So suppose $f(X)$ is stable under specialisations and $y \in \overline{f(X)}$. We need to show that $y \in f(X)$. Consider an affine open $U \ni y$. It suffices to show $y \in f(X) \cap U$.

Since $f$ is quasi-compact, $X$ is covered by a finite number of open affines $X = \cup X_i$. Therefore $y \in \overline{f(\cup X_i)}$ and since there are finitely many $X_i$-s there is an $i$ such that $y \in \overline{f(X_i)}$ (that's where we have used quasi-compactness). $\overline{f(X_i)} \cap U$ is closed hence affine, say $\mathrm{Spec}{}B$, let $X_i = \mathrm{Spec}{}A$. We have an inclusion $B \hookrightarrow A$ (since $f$ is dominant). Let $\mathfrak{p}'$ be the minimal prime ideal contained in $\mathfrak{p}_y \in \mathrm{Spec}{}B$.

Consider the localisation of $A$ at $\mathfrak{p}'$ as a $B$-module. Then $B_{\mathfrak{p}'} \hookrightarrow B_{\mathfrak{p}'} \otimes A$. But $B_{\mathfrak{p}'}$ is a field. So any proper ideal $\mathfrak{q}' \subset A \otimes B_{\mathfrak{p}'}$ has the property $\mathfrak{q}' \cap B_{\mathfrak{p}'} = (0)$.

Let $\mathfrak{q}=h^{-1}(\mathfrak{q}')$ where $h$ is the localisation map. Then $\mathfrak{q} \cap B = \mathfrak{p}'$.

☐ 

Lemma 7 (Atiyah-MacDonald, Corollary 5.22) Let $\mathcal{O}  \subset K$ be an integral local ring. Then there exists a valuation ring in $K$ that  dominates $\mathcal{O}$. It the maximal ring among local subrings of $K$ ordered by domination relation.

Proof of the valuative criterion.
Necessity. Given that $\Delta(X)$ is stable under specialisations suppose for contradiction that $h_1, h_2: \mathrm{Spec} R \rightarrow X$ are two morphisms that make the diagram commute. By the universal property of fibre product there exists a morphism $\varphi: \mathrm{Spec} R \rightarrow X \times_Y X$ such that $p_1 \circ \varphi = h_1, p_2 \circ \varphi = h_2$. Since the upper triangle of the diagram commutes, $\mathrm{Spec} K \in \mathrm{Spec} R$ is mapped by $\varphi$ to $(\mathrm{Spec} K, \mathrm{Spec} K)$. But $\varphi(\mathrm{Spec} k)$ is a specialisation of this point, hence also belongs to $\Delta(X)$. Therefore restrictions of $h_1$ and $h_2$ to $\mathrm{Spec} k$ coincide and hence $h_1 = h_2$ (by our lemma).

Sufficiency. We need to prove that $\Delta(X)$ is stable under specialisations. Let $x_1 \in \Delta(X)$ and $x_0$ be its specialisation. Then let $K = \kappa(x_1)$ and let $\mathcal{O} \subset K$ be the local ring of $x_0$ on $\overline{\{x_1\}}$ with induced reduced scheme structure. There is a valuation ring $R$, a subring of $K$, dominationg $\mathcal{O}$. We therefore have a morphism $\varphi: \mathrm{Spec}{}R \rightarrow \Delta(X)$, and $\varphi(\mathrm{Spec}{}k)=x_0$. Therefore, $x_0 \in \Delta(X)$. ☐